ERT 108 Physical Chemistry

The Second Law of Thermodynamics

by Miss Anis Atikah binti Ahmad

anisatikah@unimap.edu.my

Outline

•The Second Law of Thermodynamics•Heat Engines•Entropy•Calculation of entropy changes•Entropy, Reversibility and Irreversibility•The thermodynamics temperature scale•What is entropy?

The Second Law of Thermodynamics

•Kelvin-Plack formulation of the second law of thermodynamics:

It is impossible for a system to undergo cyclic process whose sole effects are the flow of heat into the system from a heat reservoir and the performance of an equivalent amount of work by the system on the surroundings.

The Second Law of Thermodynamics

It is impossible to build a cyclic machine that converts 100% heat into work.

Does this system violate the first law?

The Second Law of Thermodynamics

Any heat engine must eject heat into the cold reservoir

Heat Engines

•Heat engine: a device that operates in a thermodynamic cycle and does a certain amount of net positive work as a result of heat transfer from a high-temperature body to a low-temperature body.

(eg: the internal-combustion engine and the gas turbine)

Heat Engines

•The efficiency of heat engine:

• The efficiency value is less than 1, qC has a negative value and qH has a positive value.

Work output per cycle

Energy input per cycle H

C

H

CH

H

cycle

q

q

q

qq

q

w

1

Cold reservoir

Hot reservoir

Heat EnginesThe cycle for a reversible heat engine (Carnot cycle):

Heat, Work and ΔU for Reversible Carnot Cycle

Work flow in Carnot cycle

Carnot cycle

•For a complete cycle (assuming perfect gas);

•Dividing by T and integrating over Carnot cycle;

dwdqdU

PdVdq

dVVnRTdqdTCV

V

dVnR

T

dq

T

dTCV

First Law

Carnot cycle

•Thus;

V

dVnR

T

dq

T

dTCV

00

0Tdq

0 a

d

d

c

c

b

b

a T

dq

T

dq

T

dq

T

dq

T

dq

Differential of state function;independent of the path taken to reach final state.

Carnot cycle•Since bc and da are adiabatic; dq=0;

•Thus;

0 a

d

d

c

c

b

b

a T

dq

T

dq

T

dq

T

dq

T

dq0 0

0 d

c

b

a T

dq

T

dq

T

dq

0C

C

H

H

T

q

T

q

T

dq

Carnot cycleFor Carnot cycle the efficiency can be also

written as;

Where is the maximum possible efficiency for the conversion of heat to work.

rev Work output per cycle

Energy input per cycle

HCHC TTqq /Because

H

C

H

C

T

T

q

q 11

rev

Exercise 1

•Calculate the maximum work that can be done by reversible heat engine operating between 500 and 200 K if 1000 J is absorbed at 500 K

Solution•Calculate the maximum work that can be

done by reversible heat engine operating between 500 and 200 K if 1000 J is absorbed at 500 K

6.0500

200111

K

K

T

T

q

q

H

C

H

C

JJqw H 60010006.0

Entropy, S

T

dqdS rev

2

1

12 T

dqSSS rev

Closed sys, rev. process

Calculation of Entropy Changes

1. Cyclic process;

2. Reversible adiabatic process;

3. Reversible phase change at constant T & P

at constant P, , thus

0S

02

1

T

dqS rev

00S

T

qdq

TT

dqS rev

revrev

2

1

2

1

1

Hqq Prev T

S

Rev. phase change at const. T & P

Rev. adiab. proc.

Calculation of Entropy Changes

4. Reversible isothermal process

5. Constant-pressure heating with no phase change

If is constant over the temperature range, then

T

qdq

TT

dqS rev

revrev

2

1

2

1

1

dTCq PP

T

qS rev

2

1 T

dqS rev

2

1 T

dqP

dTT

CS

T

T

P2

1

PC 12ln TTCS P

Rev, isothermal proc.

Const. P, no phase change

Calculation of Entropy Changes

6. Reversible change of state of a perfect gas;

dwdUdqrev

Perfect gas

PdVdTCV

dVVnRTdTCV

2

1 T

dqS rev

2

1

2

1

dVVnRdTT

CV

2

1 1

2lnV

VnRdT

T

CV

Calculation of Entropy Changes

7. Irreversible change of state of a perfect gas;

8. Mixing of different inert perfect gases at constant T & P

Perfect gas 2

1 1

2lnV

VnRdT

T

CS V

21 SSS

bbaa VVRnVVRn lnln

bbaa xRnxRn lnln

PRTnnV ba

PRTnV aa

PRTn

PRTnnVV

a

baa

ax

1

Exercise 2 One mole of a perfect gas at 300 K is reversibly

and isothermally compressed from a volume of 25.0 L to a volume of 10.0 L. Because the water bath in the surroundings is very large, T remains essentially constant at 300 K during the process. Calculate ΔS of the system.

Exercise 2-Solution One mole of a perfect gas at 300 K is reversibly and

isothermally compressed from a volume of 25.0 L to a volume of 10.0 L. Because the water bath in the surroundings is very large, T remains essentially constant at 300 K during the process. Calculate ΔS.

• This is an isothermal process, ΔT=0, thus ΔU=0 (for perfect gas, U depends only on T. ( )

dTCdU V

0 wqU

wqrev PdV 12ln VVnRTdVV

nRT

LLKKmolJmol 2510ln300314.81 11

J310285.2

13

62.7300

10285.2

KJK

J

T

qS rev

Exercise 3

Calculate ΔS for the melting of 5.0 g of ice (heat of fusion= 79.7 cal/g) at 0°C and 1 atm.

Estimate ΔS for the reverse process

Exercise 3- Solution Calculate ΔS for the melting of 5.0 g of ice (heat

of fusion= 79.7 cal/g) at 0°C and 1 atm. Estimate ΔS for the reverse process.

• Identify type of process:▫ Phase change at constant T & P▫ At constant P, q= ΔH▫ Thus,

• Calculate ΔS;

TS

KJKcalK

ggcal

TS 1.646.1

15.273

57.79

Exercise 3- Solution Calculate ΔS for the melting of 5.0 g of ice (heat

of fusion= 79.7 cal/g) at 0°C and 1 atm. Estimate ΔS for the reverse process.

• ΔS for reverse process (freezing of 5g liquid water );

KJS 1.6

Exercise 4The specific heat capacity cP of water is nearly

constant at 100 cal/g K in the temperature range of 25°C to 50°C at 1 atm.

(a) Calculate ΔS when 100 g of water is reversibly heated from 25°C to 50°C at 1 atm.

(b) Without doing a calculation, state whether ΔS for heating 100g of water from 50°C to 75°C at 1 atm will be greater, equal to or less than ΔS for the 25°C to 50°C heating.

Exercise 4- SolutionThe specific heat capacity cP of water is nearly constant at

100 cal/g K in the temperature range of 25°C to 50°C at 1 atm.

(a) Calculate ΔS when 100 g of water is reversibly heated from 25°C to 50°C at 1 atm.

2

1 T

dqS rev

2

1 T

dqP

K

KKcal

T

TCdT

T

CP

T

T

P

298

323ln100ln

1

22

1

KJ7.33

KgcalgmcC PP 00.1100

Kcal100

Exercise 4 - SolutionThe specific heat capacity cP of water is nearly constant at 100 cal/g K

in the temperature range of 25°C to 50°C at 1 atm.

(b) Without doing a calculation, state whether ΔS for heating 100g of water from 50°C to 75°C at 1 atm will be greater, equal to or less than ΔS for the 25°C to 50°C heating.

Thus, ↑T, ↓ΔS ΔS for heating 100g of water from 50°C to 75°C at 1 atm will be

smaller than ΔS for the 25°C to 50°C heating.

TS 1

2

1 T

dqS rev

Entropy, Reversibility and Irreversibility

•Reversible Process, ΔSuniv = 0

surrsystemuniv dSdSdS

surr

rev

sys

rev

T

dq

T

dq

sys

rev

sys

rev

T

dq

T

dq

In reversible process, any heat flow btween

system & surroundings must occur with no finite

temperature difference

0

Entropy, Reversibility and Irreversibility

•Irreversible Process, ΔSuniv > 0

revrev dwdqdwdqdU

dwdwrev

When energy leaves the system as work,

dwdwrev

revrev dwdwdqdq

0 revdwdw

More work is done when a change is reversible than when it is irreversible;

Recall first law;rearranging

rearranging

•Irreversible Process, ΔSuniv > 0

Substituting into

Entropy, Reversibility and Irreversibility

revrev dwdwdqdq 0 revdwdw

0 dqdqrev

dqdqrev

T

dq

T

dqrev

T

dqdS

Clausius inequality

Dividing by T;

•Irreversible Process, ΔSuniv > 0

Suppose that the system is isolated from its surroundings, thus dq=0

Entropy, Reversibility and Irreversibility

T

dqdS 0dS

0 univsurrsys SS

•Entropy & Equilibrium

Thermodynamic equilibrium in an isolated system is reached when the system’sentropy is maximized.

Entropy, Reversibility and Irreversibility

S

Time

Equilibrium reach

S=Smax

This expression enabled Kelvin to define thermodynamic

temperature scale

The thermodynamics temperature scale

H

C

T

T1 HC TT 1

Kelvin scale is defined by using water at its triple point as the notional of hot source and defining that temperature as 273.16 K

If it is found that the efficiency of heat engine equal to 0.2, then the temperatureof cold sink is (0.8) x 273.16 K =220 K, regardless of the working substance of theengine.

rearranging

-a scale that is independent of the choice of a particular thermometric substance.

What is entropy?

•Entropy is a measure of the probability, p of the thermodynamic state

a

a a

a b b

bb

a

a a

a b b

bb

a

aa

ab

b

b

b

Partition removed

System proceed to equilibriumIrreversible mixing of perfect gas at

constant T & P

The probability that all a molecules will be in the left half & all b molecules in right half is extremely small.The most probable distribution has a and b molecules equally distributed.

What is entropy?

•Entropy is a measure of molecular disorder of a state.

Eg: In mixing two gases, the disordered (mixed state) is far more probable than the ordered (unmixed) state.

What is entropy?

• Entropy is related to the distribution or spread of energy among the available molecular energy levels.

• The greater the number of energy levels, the larger the entropy is.

• Increasing the system’s energy (eg:by heating) will increase its entropy because it allows higher energy levels to be significantly occupied

• Increasing the volume of a system at constant energy also allows more energy level to be occupied.

What is entropy?

•Boltzmann made link btween distribution of molecules over energy levels and the entropy;

Where k= 1.381 x 10-23 JK-1

W= probability/ways in which the molecules of a system can be

arranged while keeping the energy constant

WkS ln

Exercise

•True or false?▫ΔSuni for a reversible process in a

closed system must be zero▫ΔS for a reversible process in a closed

system must be zero▫For a closed system, equilibrium has

been reached when S has been maximized.

•What is ΔSuni for each steps of a Carnot cycle?